Calculus: Early Transcendentals, Chapter 4, 4.1, Section 4.1, Problem 35

You need to find the critical points of the function, hence, you need to evaluate the solutions to the equation g'(y) = 0.
You need to evaluate the first derivative, using the quotient rule:
g'(y)= ((y-1)'(y^2-y+1) - (y-1)(y^2-y+1)')/((y^2-y+1)^2)
g'(y)= (y^2-y+1 - (y-1)(2y-1))/((y^2-y+1)^2)
g'(y)= (y^2-y+1 - 2y^2 + 3y - 1))/((y^2-y+1)^2)
g'(y)= (-y^2+2y)/((y^2-y+1)^2)
You need to solve for g'(y) = 0, such that:
-y^2+2y = 0 => y^2 - 2y = 0
Factoring out y yields:
y(y - 2) = 0 => y = 0
y - 2 = 0 => y = 2
Hence, evaluating the critical numbers of the function for g'(y) = 0, yields y = 0 and y = 2.